Chapter 6 : Probabilities, Combinations and Permutations

Dr Roger Nix (Queen Mary, University of London) - 6.1

CHAPTER 6: PROBABILITY, COMBI�ATIO�S & PERMUTATIO�S

Probability

If an event can occur in n ways (i.e. there are n possible outcomes) and a particular result can occur

in m ways, then the probability of the particular result occurring is m/n .

Example

Spinning a coin gives rise to 2 possible outcomes: H (Heads) or T (Tails)

H occurs in 1 way ; therefore the probability of a H result is ½

Note:

1. the sum of the probabilities for all possible results is unity: in the above example, the probability

of T occurring is also ½ and the probability of obtaining either an H or a T (i.e. encompassing all

possible results) is therefore equal to (½ +½) = 1

2. probabilities may be expressed as fractions, in decimal format or as percentages; so the following

probabilities are all equivalent - ½ , 0.5 , 50%.

Example

A standard dice has six faces (with faces numbered 1 …. 6) and rolling it gives rise to 6 possible

outcomes. A 3 occurs in 1 way ; therefore the probability of throwing a 3 is 1/6

Permutations and Combinations

If we have n different items, then the number of ways of arranging them in a row is given by the

factorial n!

Proof by Example

Consider how many arrangements are possible for the 26 letters of the alphabet. The first letter

could be chosen in 26 different ways - leaving 25. Thus 25 choices are possible for the second

letter. The third choice is then made from the 24 remaining letters and so on. The number of

different arrangements possible for just the first two choices is thus 26 × 25 ; and when the third

is included, 26 × 25 × 24. This process can be continued until all the letters of the alphabet are

used up and the number of different possible arrangements is 26 × 25 × 24 × 23 × ......... × 4 × 3

× 2 × 1 - such a multiplication series is called a factorial and in this case the number of

arrangements would be written as 26!

Example

With three different letters, ABC, then n = 3 and n! = 6 , corresponding to the six possible

arrangements which are: ABC, ACB, BAC, BCA, CAB, CBA


Selections

We are often concerned with considering selections of objects, rather than all the objects. In this

instance it is important to distinguish between two different possible situations; one in which the order

of the objects is important and one in which the order is not.

Example

It is possible to select two letters from the letters ABC in 3 ways: AB, AC, BC

Each selection is called a combination

However,

If the order of the selection matters then there are 6 ways: AB, BA, AC, CA, BC, CB

Each arrangement is called a permutation

In this case, therefore there are 3 combinations and 6 permutations of the selected two letters.

In general, provided that all the objects are different (dissimilar), then

• the number of combinations of r objects from n objects is: n

rCn

n r r=

−

!

( )! !

• the number of permutations of r objects from n objects is: n

rPn

n r=

−

!

( )!

For the example given above, the number of combinations of two letters chosen from three is

3

2

3

1 23C = =

!

! ! , whilst the number of permutations of two from three is 3 2

3

16P = =

!

!

The difference between the two formulae is a factor of r! - this is the number of ways of arranging r

dissimilar objects. For two different letters, r = 2 and r! = 2 , i.e. for a pair of letters, then each letter

may come first or last, giving rise to two permutations for each combination.

Example

How many different arrangements can be made by taking 5 letters of the word numbers ?

Number of permutations = 7 5

7

22520P = =

!

!

Example

How many different arrangements can be made by taking all the letters of the word numbers ?

Number of permutations = 7 7

7

05040P = =

!

!


Mass Spectroscopy - Isotopic Patterns

Example 1 : Cl and Cl2

Chlorine has two stable isotopes, with integer masses of 35 and 37. Their relative abundance is about

3 : 1. This can be expressed as ″ the chance of encountering a Cl atom of mass 35 in a collection of

free chlorine atoms is 75% ″ or the probability of any particular Cl atom having mass 35 is 0.75

This means that a mass spectrum of chlorine atoms shows two peaks at masses 35 and 37, the former

three time as intense as the latter.

What happens in the chlorine molecule (Cl2 )?

Clearly several combinations of isotopes are possible, (i) both atoms are of mass 35, giving 35Cl2 ,

(ii) one 35 and one 37, giving 35Cl

37Cl, or (iii) both are of mass 37, giving

37Cl2 . These possible

combinations give rise to three peaks in the mass spectrum, corresponding to Cl2+ ions with relative

masses of 70, 72, and 74.

There are several ways of tackling the problem of identifying the exact probability of each molecular

mass occurring. We will start with the least sophisticated approach.

Approach 1

If we choose the first atom in the molecule to be 35Cl then this choice has a probability of 0.75; this

atom may then be associated with either another 35Cl atom, again with a probability of 0.75, or with a

37Cl atom, with a probability of 0.25 .

The pattern of all possible combinations, taking account of these relative probabilities is shown in

Table 1. For this table we have chosen to work with four chlorine atoms in the first instance, since the

relative abundances then indicate that on average three will be 35 and one 37. Each of these atoms

can then be associated with four other atoms, and again three will be 35 and one 37, giving the sixteen

possible C12 molecules shown in the table.

Table 1 : Possible Ways of Combining Chlorine Isotopes

First Cl 35 35 35 37

Second Cl 35 35 35 37 35 35 35 37 35 35 35 37 35 35 35 37

Total mass 70 70 70 72 70 70 70 72 70 70 70 72 72 72 72 74

Total number of molecules = 16

No. of molecules of mass 70 = 9 ⇒ Relative probability = 9/16 = 0.5625




Cross-checking:

Total number of molecules = 9 + 6 + 1 = 16

Total probability of all outcomes = 0.5625 + 0.3750 + 0.0625 = 1.0000

Note that the total number of possible isotopic arrangements in the final molecule arises by

multiplying the number of choices for the first atom by the number of choices for the second. In this

case four for the first times four for the second i.e. sixteen.

If the only way in which these molecules can be distinguished is by mass then the relative probabilities

will be as shown on the right hand side of the lower part of the table – of the sixteen arrangements,

nine correspond to a total molecular mass of 70, six correspond to a total mass of 72 and one

corresponds to a total mass of 74. The mass spectrum will therefore appear as shown below:

Approach 2

The same result could be achieved, with considerably less effort, if the isotopic probability were used

directly. The four possible isotopic arrangements are listed in the table below, and their probabilities

evaluated. It is very important to note that the probability of a particular combination or permutation

is related to the product of the probabilities.

Table 2 : Possible Permutations of two Chlorine Isotopes with their Probabilities

1st 2nd Probability Product Probability of Permutation Total Mass

35 35 0.75 × 0.75 0.5625 70

35 37 0.75 × 0.75 0.1875 72

37 35 0.25 × 0.75 0.1875 72

37 37 0.25 × 0.25 0.0625 74

Sum 1.0000

The resulting conclusion is the same as that derived from the more complicated Table 1 - the

probability of the molecule having a total mass of 70 is 0.5625, a total mass of 72 is 0.3750

(0.1875+0.1875), and a total mass of 74 is 0.0625

Intensity

70 72 74 Mass

Intensity ratio = 9 : 6 : 1


Approach 3

A final simplification results from taking direct account of how many permutations give rise to a

particular molecular mass. The final probability is now determined directly from multiplying the

product of the individual probabilities by the number of permutations of the isotopes giving the

required total mass.

Table 3 : Possible Permutations of two Chlorine Isotopes with their Probabilities

Total Mass Required Isotopes Probability Product No. of Permutations Overall Probability

70 Two 35 0.75 × 0.75 = 0.5625 1 0.5625

72 One 35; One 37 0.75 × 0.25 = 0.1875 2 0.3750

74 Two 37 0.25 × 0.25 = 0.0625 1 0.0625

Example 2 : CHCl3

Let’s now consider a slightly more complicated example, that of chloroform CHCl3 . We will assume

for simplicity that C is only present as mass 12 and H is only present as mass 1 in the parent ion,

CHCl3+- the molecular mass is therefore that due to the three chlorine atoms plus 13. The pattern and

masses of the peaks derived from this ion can be calculated as shown below.


118 Three 35 0.75 × 0.75 × 0.75 1 0.4219

120 Two 35; One 37 0.75 × 0.75 × 0.25 3 0.4219

122 One 35; Two 37 0.75 × 0.25 × 0.25 3 0.1406

124 Three 37 0.25 × 0.25 × 0.25 1 0.0156

Sum 1.0000

Note:

1. For the 120 amu ion, which must contain two 35 and one 37 isotopes, there are three ways in

which the isotopes can be arranged in order (35-35-37, 35-37-35 and 37-35-35). Similarly there

are also three permutations for the 122 amu ion.

2. The final probabilities have been added up and the sum is equal to unity. This is quite general and

is a very useful check which should always carried out at this stage in your calculations.

If the sum of the probabilities isn’t equal to one then you have made an error.

Example 3 : BrCl

More complicated situations arise if many atoms in a molecule have a range of possible isotopes, but

no new concepts arise. Thus the parent ion of bromine monochloride, Br Cl+ , will show four peaks in

the its mass spectrum since both bromine and chlorine have two isotopes. The isotopes and their

relative probabilities are :


79Br 0.50

81Br 0.50

35Cl 0.75

37Cl 0.25

For more complex problems such as this one the simplest procedure is to consider the isotopic

arrangements associated with each element individually and then to combine the probabilities

calculated for groupings of like atoms.

In the case of BrCl+ this is very simple:

Br C1 Mass Br Proby × Perms

Cl Proby × Perms

Probability products

35 114 0.5 × 1 = 0.5 0.75 × 1 = 0.75 0.5 × 0.75 = 0.375

79

37 116 0.5 × 1 = 0.5 0.25 × 1 = 0.25 0.5 × 0.25 = 0.125

35 116 0.5 × 1 = 0.5 0.75 × 1 = 0.75 0.5 × 0.75 = 0.375

81

37 118 0.5 × 1 = 0.5 0.25 × 1 = 0.25 0.5 × 0.25 = 0.125

Sum 1.000

Note that mass 116 arises from both 79Br

37Cl and

81Br

35Cl and in a mass spectrometer that can only

resolve unit atomic masses just one peak will result – this will have an intensity reflecting the sum of

the probabilities (i.e. 0.500).

This situation is however fundamentally different from that observed for different permutations (e.g. 35Cl

37Cl ,

37Cl

35Cl ) since in the 116 peak of BrCl two different elements are involved. The truth of

this can be seen if a very high resolution spectrometer is used in which case the peaks would be

resolved. The exact masses of the different isotopes then need to be considered:

For 81Br

35Cl,

81Br 80.9163

35Cl 34.9802

Total molecular mass = 115.8965

For 79 Br

37Cl,

79Br 78.9183

37Cl 36.9776

Total molecular mass = 115.8859

Such a small difference can be observed experimentally, whereas the mass of 35Cl

37Cl will always be

the same as that of 37Cl

35Cl no matter how accurate the spectrometer.

The spectrum would now look rather different on spectrometers of different resolution (although this

is exaggerated in the diagram below):


(i) unit mass resolution (ii) very high resolution

Example 4 : B2Cl2

A slightly more complicated example is that of the ion B2C12+. Here the relevant atomic isotopic

masses and probabilities are:

10B 0.2

11B 0.8

35Cl 0.75

37Cl 0.25

(i) First considering the two B atoms


20 Two 10B 0.2 × 0.2 = 0.04 1 0.04

21 One 10B; One

11B 0.2 × 0.8 = 0.16 2 0.32

22 Two 11B 0.8 × 0.8 = 0.64 1 0.64

Sum 1.00

(ii) Secondly considering the two Cl atoms


70 Two 35 0.75 × 0.75 = 0.5625 1 0.5625

72 One 35; One 37 0.75 × 0.25 = 0.1875 2 0.3750

74 Two 37 0.25 × 0.25 = 0.0625 1 0.0625

Sum 1.0000

(iii) Combine results for the two sub-units.

For the ion B2 Cl2+ the probability for the B2 group and the Cl2 group must be multiplied

together thus:

Intensity

114 116 118 Mass 114 116 118 Mass


B2 C12 Mass of B2Cl2 B2 Proby Cl2 Prob

y Probability Product Probability

70 90 0.04 0.5625 0.04 × 0.5625 0.0225

20 72 92 0.04 0.3750 0.04 × 0.3750 0.0150

74 94 0.04 0.0625 0.04 × 0.0625 0.0025

70 91 0.32 0.5625 0.32 × 0.5625 0.1800

21 72 93 0.32 0.3750 0.32 × 0.3750 0.1200

74 95 0.32 0.0625 0.32 × 0.0625 0.0200

70 92 0.64 0.5625 0.64 × 0.5625 0.3600

22 72 94 0.64 0.3750 0.64 × 0.3750 0.2400

74 96 0.64 0.0625 0.64 × 0.0625 0.0400

Sum 1.0000

A simple mass spectrum (at unit mass resolution) would therefore contain seven peaks due to the

molecular ion with the following relative intensities.

Mass of B2Cl2 Probability

90 0.0225

91 0.1800

92 0.3750

93 0.1200

94 0.2425

95 0.0200

96 0.0400

but the 92 and 94 peaks would show a further splitting if studied at much higher resolution.

Example 5 : MCl6

In every example considered so far the number of permutations associated with a certain selection of

isotopes has been easily obtained by inspection. However, for more complex molecules we need to

have a more systematic procedure.

For example in hexachloro complexes (MCl6 ) the Cl6 unit can have seven different masses arising

from the following combinations of isotopes:

1. all 35

2. five 35 + one 37

3. four 35 + two 37

4. three 35 + three 37

5. two 35 + four 37

6. one 35 + five 37

7. all 37


In each case the number of possible arrangements must be calculated; for (1) and (7) it is obviously

just one, but what about the rest ?

General Case : would be n possible atoms divided into groups of r1 , r2 , ..... , ri similar atoms , such

that r1 + r2 + r3 ....+ ri = n (i.e. nri =∑ ).

If it is first assumed that the n atoms are all different isotopes then the total number of possible

arrangements would (as discussed earlier) be n! . If, however, some atoms are identical then a

rearrangement of these does not generate a new permutation. If r are identical then there are r!

hypothetical arrangements of these r identical isotopes. The original total number of arrangements

must be reduced by this number. More generally the original number of arrangements must be

reduced by the number of arrangements of identical atoms in each group, i.e. by r1! and by r2 ! etc.

i.e. in general the final number of distinct arrangements would be

! ! ! !

!

321 ir.....rrr

n

×××

For the particular case of Cl6 considered above ( i.e. with n = 6 ):

No. of 35Cl atoms No. of

37Cl atoms N

o. of Permutations

( r1 ) ( r2 )

6 0 !0 !6

!6

× = 1

5 1 !1 !5

!6

× = 6

4 2 !2 !4

!6

× = 15

3 3 !3 !3

!6

× = 20

2 4 !4 !2

!6

× = 15

1 5 !5 !1

!6

× = 6

0 6 !0 !6

!6

× = 1

These numbers must now be multiplied by the appropriate isotopic probabilities to give the overall

probability of finding Cl6 unit of a particular mass.

For example, the probability of a Cl6 unit having mass 214 ( 35Cl4

37Cl2 ) is given by :

P = 15 × (0.75 × 0.75 × 0.75 × 0.75) × (0.25 × 0.25) = 15 × (0.75)4 × (0.25)2 = 0.30


�uclear Magnetic Resonance : Spin -spin coupling

Protons can be regarded as spinning positive charges: they have a magnetic moment associated with

this ′motion′. The magnetic moment may either be aligned with or opposed to an applied magnetic

field giving two possible energy states of the proton. If the appropriate amount of energy is supplied,

as radiation of a particular frequency, then a transition can be effected from the more stable to the less

stable state and the photons of that frequency are absorbed. This is the basic principle of nuclear

magnetic resonance (NMR) spectroscopy. The protons of hydrogen atoms in a molecule do not see an

applied magnetic field in isolation, but rather the applied field modified by the local magnetic field of

that part of the molecule in which they are sited. Thus protons in different parts of a molecule will

absorb radiation at slightly different frequencies, and so they can be identified. An NMR spectrum

will thus show many peaks corresponding to the different types of hydrogen present in a molecule e.g.

those of a CH3 - group, CH2 - group, C(aromatic)–H etc.

A closer examination of such peaks will often reveal a fine structure which is due to ′spin-spin′

interaction, i.e. the effect of the magnetic state of one group of protons on another group. For example

with the ethyl group CH3-CH2- the methylene (CH2 ) peak is split into four sub-peaks and the methyl

(CH3 ) peak into three. The spins of the three methyl hydrogen atoms can each be ′up′ or ′down′ ; for

simplicity the two possible states can be designated α or β . Overall the three methyl hydrogens may

exhibit four different possible spin arrangements which are:

1. all α 2. two α , one β

3. one α , two β 4. all β

The local magnetic field seen by the protons of the methylene is modified, slightly, by the magnetic

state of the adjacent methyl group. Furthermore as four possible magnetic states are possible for the

methyl group, the methylene resonance can take place at four slightly different frequencies hence the

observed splitting of the methylene peak. However the probabilities of finding the four spin

configurations (1-4) are not equal. Since the probabilities of the two possible spin states, α and β , are

equal, the probability of any particular spin configuration for the methyl group depends simply upon

the number of permutations giving rise to the configuration. They will be.

Spin α Spin β No. of Permutations

3 0 !0 !3

!3

× = 1

2 1 !1 !2

!3

× = 3

1 2 !2 !1

!3

× = 3

0 3 !0 !3

!3

× = 1

Thus not only will the methylene peak be split into four sub-peaks but their relative intensities will be

1 : 3 : 3 : 1. Similar arguments show that the methylene protons will cause the methyl peak to be split

into three, with relative sub-peak intensities of 1 : 2 : 1.


Binomial Expansion

In Section 1 (1.10) we looked at the indexing of bracketed expressions. We can now obtain a general

expression for (x + y) n, where n is any positive integer, by considering how the above expressions

are obtained during the expansion of the brackets.

For n = 2, (i.e. (x + y)2 ) there is only one way of taking x from each of the two brackets to give an xx

or x2 term. Similarly there is only one way of taking one y from each bracket, giving yy or y

2. But

there are two ways of taking an x from one bracket and a y from the other bracket, giving xy and

yx , i.e. giving 2xy in total. Hence (x + y)2 = 1x

2 + 2xy + 1y

2.

For n = 3, (i.e. (x + y)3 ) there is again one way of taking x from each of the three brackets, giving

xxx or x3, and only on way of taking y from each of the three brackets, giving yyy or y

3. But there

are three ways of selecting two x’s and one y, giving xxy , xyx and yxx , i.e. 3x2y. Similarly, for one x

and two y’s, there are three ways giving xyy, yxy and yyx , i.e. 3xy2 . Hence (x + y)

3 = 1x

3 + 3x

2y +

3xy2 + 1y

3.

From the pattern of coefficients which is emerging it is possible to recognise that the general

expression will be of the form :

(x + y) n =

nC0 x

n +

nC1 x

n–1y +

nC2 x

n–2y2 + ........... +

nCr x

n–ry r + .... +

nCn y

n,

where the binomial coeffecients nC0 .....

nCr .....

nCn represent the number of ways in which the

corresponding terms x n ....... x

n–ry r........ y

n can be formed in the expansion. These coefficients may

be evaluated using the expression given on page 6.2, i.e.

n

rCn

n r r=

−

!

( )! !

or they may be obtained from Pascal's triangle, as described on page 1.10 .

Documents

Chapter 6 : Probabilities, Combinations and Permutations